Question Number 32334 by abdo imad last updated on 23/Mar/18
$${p}\:{is}\:{apolynomial}\:{having}\:{n}\:{roots}\:\:\left({x}_{{i}} \right)\:{with}\:{x}_{{i}} \neq{xj}\:{for} \\ $$$${i}\neq{j}\:\:{calculate}\:\:\sum_{{k}=\mathrm{1}} ^{{n}} \:\:\:\frac{\mathrm{1}}{\mathrm{1}−{x}_{{k}} }\:. \\ $$
Commented by abdo imad last updated on 01/Apr/18
$${p}\left({x}\right)=\lambda\:\prod_{{k}=\mathrm{1}} ^{{n}} \:\left({x}−{x}_{{i}} \right)\:\Rightarrow{p}^{'} \left({x}\right)=\lambda\:\sum_{{k}=\mathrm{1}} ^{{n}} \:\prod_{{p}=\mathrm{1}_{{p}\neq{k}} } ^{{n}} \:\left({x}−{x}_{{p}} \right) \\ $$$$\Rightarrow\:\frac{{p}^{'} \left({x}\right)}{{p}\left({x}\right)}\:=\:\sum_{{k}=\mathrm{1}} ^{{n}} \:\:\frac{\mathrm{1}}{{x}−{x}_{{k}} }\:\Rightarrow\:\sum_{{k}=\mathrm{1}} ^{{n}} \:\:\frac{\mathrm{1}}{\mathrm{1}−{x}_{{k}} }\:=\frac{{p}^{'} \left(\mathrm{1}\right)}{{p}\left(\mathrm{1}\right)}\:\:{with}\:{p}\left(\mathrm{1}\right)\neq\mathrm{0} \\ $$$$ \\ $$